Chemistry · D And F Block Elements · NEET
Two reasons that NCERT states clearly. (1) Variable oxidation states: the metal can gain or give electrons easily by switching between states like +2 and +3. This gives the reactant an easy, low-energy path. (2) Complex formation with ligands: the metal can bind the reactant molecules close together on its surface, so they react more easily. Extra point: transition metals also have large surface area when finely divided, which adsorbs (sticks) reactant gases and helps them react. So the answer NEET wants is 'variable oxidation states and formation of complexes'.
NCERT gives the reaction between iodide (I-) and persulphate (S2O8^2-), which is slow on its own. Fe3+ acts as a catalyst: 2Fe3+ + 2I- -> 2Fe2+ + I2, then 2Fe2+ + S2O8^2- -> 2Fe3+ + 2SO4^2-. Iron switches between +3 and +2, carrying electrons between the two reactants. The iron is regenerated at the end, so a small amount keeps working. This switching is only possible because transition metals have variable oxidation states.
Learn these four cold, because NEET repeats them: Finely divided IRON (Fe) with a promoter is used in the Haber process to make ammonia (N2 + 3H2 -> 2NH3). V2O5 (vanadium pentoxide) is used in the Contact process to oxidise SO2 to SO3 for making H2SO4. PdCl2 catalyses the Wacker oxidation of ethyne/alkene to ethanal (acetaldehyde). TiCl4 + Al(CH3)3 is the Ziegler-Natta catalyst for polymerisation of ethylene. Nickel (Ni) is used for hydrogenation of oils and Ni complexes for polymerisation of alkynes.
Iron (Fe) = Haber process = making AMMONIA (NH3). V2O5 = Contact process = making SULPHURIC ACID (H2SO4) by oxidising SO2. Memory trick: 'F' in Fe matches 'FertiliseR' (ammonia makes fertiliser); 'V' in V2O5 has a shape like the 'A' in Acid (sulphuric acid). Do not swap them - NEET loves to test exactly this match.
No. A catalyst speeds up the reaction but is regenerated (comes back) at the end, so it is not used up. In the Fe3+ example, Fe3+ becomes Fe2+ and then returns to Fe3+. This is why a tiny amount of catalyst can process a large amount of reactant. The catalyst also does not change the equilibrium constant K; it only helps the reaction reach equilibrium faster.
s-block metals like Na and Ca have only ONE stable oxidation state (Na is +1, Ca is +2). They cannot easily switch states to carry electrons, and they do not form stable complexes with empty d-orbitals. Transition metals have partly filled d-orbitals, so they show variable oxidation states AND form complexes - both needed for catalysis. That is why catalysis is a special d-block property.
Match List I with List II. List I: (a) V2O5 (b) Fe (c) PdCl2 (d) Ni complex. List II: (i) Preparation of ammonia from N2/H2 mixture (ii) Polymerisation of alkynes (iii) Preparation of H2SO4 (oxidation of SO2) (iv) Oxidation of ethyne to ethanal.
Match the catalyst with the process. Catalyst: (i) V2O5 (ii) TiCl4 + Al(CH3)3 (iii) PdCl2 (iv) Nickel complexes. Process: (a) oxidation of ethyne to ethanal (b) polymerisation of alkynes (c) oxidation of SO2 in H2SO4 manufacture (d) polymerisation of ethylene.
Identify the incorrect statement.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Variable (multiple) oxidation states and the ability to form complexes with ligands. Both give reactant molecules an easier, lower-energy path to react. Finely divided metals also adsorb reactants on their large surface, which helps too.
Finely divided iron (Fe) with a promoter (like molybdenum). It makes ammonia: N2 + 3H2 -> 2NH3.
Vanadium pentoxide, V2O5. It oxidises SO2 to SO3 in the manufacture of sulphuric acid, H2SO4.
A mixture of TiCl4 and Al(CH3)3 (triethyl/trimethyl aluminium). It is used to polymerise ethylene into polythene.
No. A catalyst only speeds up how fast equilibrium is reached. It does not change the value of the equilibrium constant K or the position of equilibrium, and it is not consumed.
